Question

Find dy/dx and d2y/dx2. x = cos(2t), y = cos(t), 0 < t < π

Answer 1

`(\mathrm dy)/(\mathrm dx)=((\mathrm dy)/(\mathrm dt))/((\mathrm dx)/(\mathrm dt))`


`y=\cos t\implies(\mathrm dy)/(\mathrm dt)=-\sin t`

`x=\cos2t\implies(\mathrm dx)/(\mathrm dt)=-2\sin2t=-4\sin t\cos t`

`\implies(\mathrm dy)/(\mathrm dx)=(-\sin t)/(-4\sin t\cos t)=\frac14\sec t`

Let
`z=(\mathrm dy)/(\mathrm dx)`. Then


`(\mathrm d^2y)/(\mathrm dx^2)=(\mathrm dz)/(\mathrm dx)=(\mathrm dz)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)=((\mathrm dz)/(\mathrm dt))/((\mathrm dx)/(\mathrm dt))`

`\implies(\mathrm d^2y)/(\mathrm dx^2)=((\mathrm d)/(\mathrm dt)\left[\frac14\sec t\right])/(-4\sin t\cos t)=(\frac14\sec t\tan t)/(-4\sin t\cos t)=-\frac1{16}\sec^3t`